Topic 5. Chemical Energetics
Created by S. Colgan;
Modified by K. Slater
Resources:
http://lincoln.pps.k12.or.us/lscheffler/Energetics.ppt#269,36,Sta
ndard Enthalpy Changes
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IB Topic 5: Energetics
5.1: Measuring Energy Changes
Understandings
 Heat is a form of energy.
 Temperature is a measure of the average kinetic energy
of the particles.
 Total energy is conserved in chemical reactions.
 Chemical reactions that involve transfer of heat between
the system and the surroundings are described as
endothermic or exothermic.
 The enthalpy change (∆H) for chemical reactions is
indicated in kJ mol-1.
 ∆H values are usually expressed under standard
conditions, given by ∆H°, including standard states.
2
Chemical Reactions
In a chemical reaction
 Chemical bonds are broken
 Atoms are rearranged
 New chemical bonds are formed
 These processes always involve
energy changes
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3
Laws of Conservation – Review
The Law of Conservation of…
Matter
 Mass
 Energy

Write out the balanced equation for the
combustion of ethene, C2H4(g).
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4
Energy Changes

Breaking chemical bonds requires energy
 The


system requires energy from the surroundings
(leave this blank for later)
Forming chemical bonds releases energy
 The

system gives off energy to the surroundings
(leave this blank for later)
 Let’s
calculate the energy change in the
combustion of ethene.
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5
Temperature and Heat
Temperature: a measure of the
average kinetic energy of a body
 Heat: energy that is transferred from
one object to another due to a
difference in temperature


Heat is always transferred from objects at a
higher temperature (more KE) to those at a
lower temperature (lower KE)

http://ed.ted.com/lessons/all-of-the-energy-in-the-universeis-george-zaidan-and-charles-morton#watch
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5.1.1 Define the terms exothermic reaction,
endothermic reaction and standard enthalpy
change of reaction (ΔHo).
Enthalpy (ΔH): The heat content of a
system
When heat is added to the system from the
surroundings, enthalpy increases  ΔH is
positive
 When heat is released from the system to the
surroundings, enthalpy decreases  ΔH is
negative

7
5.1.1 Define the terms exothermic reaction,
endothermic reaction and standard enthalpy
change of reaction (ΔHo).
Exothermic Reaction: Energy transfer from
the system to its surroundings
ΔH is negative
 Releases heat / feels warm
 Products have less energy than the reactants

Endothermic Reaction: Energy transfer
from the surroundings to the system
ΔH is positive
 Absorbs heat
 Products have more energy than the reactants
8

Origami Definitions
Heat
 Enthalpy
 Endothermic
 Exothermic

9
5.1.3 Apply the relationship between temperature
change, enthalpy change and the classification of a
reaction as endothermic or exothermic.
Exothermic
Heat flows out of the
system
Surroundings heat up
ΔH is negative
C8H18+ 12½O2  8CO2 +
9H2O ΔH = -5512 kJ mol-1
H2 + ½O2  H2O
ΔH = -286 kJ mol-1
Endothermic
Heat flows into the system
Surroundings cool down
ΔH is positive
H2O(s)  H2O(l)
ΔH = +6.01 kJ mol-1
½N2 + O2  NO2
ΔH = +33.9 kJ mol-1
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Energy Changes

Forming new chemical bonds releases
energy
 Exothermic

Breaking chemical bonds requires energy
 Endothermic
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Enthalpy Level Diagrams
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5.1.4 Deduce, from an enthalpy level diagram, the
relative stabilities of reactants and products, and the
sign of enthalpy change for the reaction.
Exothermic Reactions
Products more stable than reactants
(lower energy).
Since the products have less
energy than the reactants, the ΔH
value is negative.
Endothermic Reactions
Products less stable than reactants
(higher energy)
Since the products have more
energy than the reactants, the ΔH
value is positive.
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Exothermic and Endothermic
Processes

Exothermic processes release energy
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4H2O (g)
+ 2043 kJ

Endothermic processes absorb energy
C(s) + H2O (g) +113 kJ  CO(g) + H2 (g)
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Energy Changes in endothermic
and exothermic processes
In an
endothermic
reaction there is
more energy
required to
break bonds
than is released
when bonds are
formed.
The opposite is
true in an
exothermic
reaction.
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5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic
If…
Bond enthalpy reactants > Bond enthalpy products
 Reaction is endothermic
Since the amount of energy required to break the
bonds in the reactants is greater than the
amount of energy released when bonds are
formed in the products, the reaction is
endothermic.
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5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic
If…
Bond enthalpy reactants < Bond enthalpy products
 Reaction is exothermic
Since the amount of energy required to break the
bonds in the reactants is less than the amount
of energy released when bonds are formed in
the products, the reaction is exothermic.
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Energy Changes
If we know values for bond strength (bond
enthalpy), then we can calculate the heat
energy exchanged with the surroundings
when a reaction happens under standard
conditions (NOT STP) = Standard
enthalpy of reaction
 Enthalpy changes of reactions are the
result of bonds breaking and new bonds
being formed.

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Bond Enthalpy
Remember…

Breaking bonds requires energy

Forming new bonds releases energy
Bond enthalpy is the energy required to break one mole
of a certain type of bond in the gaseous state
averaged across a variety of compounds.
FYI: Bond enthalpies for unlike atoms will be affected by
surrounding bonds and will be slightly different in
different compounds so average bond enthalpies
are used.
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5.3.2 Using bond enthalpy to determine
enthalpy change of a reaction
The average bond enthalpies for several types of
chemical bonds are shown in the table below:
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Methods of Calculating Enthalpy
There are four ways of calculating ΔH for a
reaction: by using…
1. bond enthalpies
2. experimental data to calculate q

3.
4.
then dividing by 1000 and moles to get kJ
mol-1
enthalpy of formation values
Hess’s Law
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5.3.2 Using bond enthalpy to determine
enthalpy change of a reaction
 H
=
∑ (energy required
to break bonds)
OR
 H
–
∑(energy released
when bonds are formed)
–
∑(bond enthalpy
of products)
=
∑ (bond enthalpy
of reactants)
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5.3.2 Using bond enthalpy to determine
enthalpy change of a reaction
H =
∑ (bond enthalpy
of reactants)
–
∑(bond enthalpy
of products)
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5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic
Use the following average bond enthalpies (kJ mol-1) to determine the
heat of reaction for 3F2 + NH3  3HF + NF3
F-F = 158; N-H = 388; H-F = 562; N-F = 272
Energy in (kJ mol-1)
3F-F is 3(158) + 3N-H is 3(388) = 1638 kJ
Energy out (kJ mol-1)
3H-F is 3(562) + 3N-F is 3(272) = 2502 kJ
Since Energy in < Energy out, the reaction is exothermic
Heat of reaction is -864 kJ mol-1
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5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic
Using the Bond Enthalpy Table, determine the heat of reaction for:
CO(g) + 2H2(g)  CH3OH
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5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic
Using the Bond Enthalpy Table, determine the heat of reaction for:
CO(g) + 2H2(g)  CH3OH
Energy in (kJ)
1CO (triple bond) 1074 + 2H-H is 2(436) = 1946 kJ mol-1
Energy out (kJ)
3C-H is 3(413) + 1C-O is 358 + 1 O-H is 464 = 2061 kJ mol-1
Hrxn = 1946 – 2061 = -115 kJ mol-1
Since Energy in < Energy out, reaction is exothermic
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5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic
Using the Bond Enthalpy Table, determine the heat of reaction for:
CO(g) + 2H2(g)  CH3OH
Energy in (kJ)
1CO (triple bond) 1074 + 2H-H is 2(436) = 1946 kJ mol-1
Energy out (kJ)
3C-H is 3(413) + 1C-O is 358 + 1 O-H is 464 = 2061 kJ mol-1
Hrxn = 1946 – 2061 = -115 kJ mol-1
Since Energy in < Energy out, reaction is exothermic
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Bond Enthalpy Calculations
Calculate the enthalpy change for the
reaction N2 + 3 H2  2 NH3
Bonds broken
1 N=N:
3 H-H:
= 945
3(435) = 1305
Total = 2250 kJ
Bonds formed
2x3 = 6 N-H:
6 (390) = - 2340 kJ
Net enthalpy change
= + 2250 - 2340 = - 90 kJ
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5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic
In the 1960s NASA considered the relative merits of using hydrogen
and oxygen compared with hydrogen and fluorine as rocket fuels.
Assuming all the reactants and products are in the gaseous state,
use bond enthalpies to calculate the enthalpy change of reaction
(in kJ mol-1 of product) for both fuels. As mass is more important
than amount in the choice of rocket fuels, which reaction would
give more energy per kilogram of fuel?
Bond enthalpies (kJ mol-1):
H-H: 435; O O: 496; H-O: 464 kJ; F-F: 158; H-F: 562
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5.1.1 Define the terms exothermic reaction,
endothermic reaction and standard enthalpy
change of reaction (ΔHo).
Standard Enthalpy Change of Reaction (∆H): The heat
energy exchanged with the surroundings when a
reaction happens under standard conditions (NOT STP…
see below).
Since the enthalpy change for any given reaction will vary
with the conditions, ΔH are measured under standard
conditions:




pressure = 101.3 kPa
temperature = 25ºC = 298 K
Concentrations of 1 mol dm-3
The most thermodynamically stable allotrope (which in the case of
carbon is graphite)
Only ΔH can be measured, not H for the initial or final
state of a system.
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5.1.1 Define the terms exothermic reaction,
endothermic reaction and standard enthalpy
change of reaction (ΔHo).
Pseudonyms (other names) for H
 Heat of Reaction: Hrxn heat produced in a chemical reaction
 Heat of Combustion: Hcomb heat produced by a combustion
reaction
 Heat of Neutralization: heat produced in a neutralization reaction
(when an acid and base are mixed to get water, pH = 7)
 Heat of solution: Hsol heat produced by when something dissolves
 Heat of Fusion: Hfus heat produced when something melts
 Heat of Vaporization: Hvap heat produced when something
evaporates
 Heat of Sublimation: Hsub heat produced when something sublimes
 Heat of formation: Hf change in enthalpy that accompanies the
formation of 1 mole of compound from it’s elements (this has special
uses in chemistry…)
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Combustion and neutralization
reactions are exothermic processes
Combustion
Exothermic reaction
General Combustion Reaction Formula:
Compound (usually hydrocarbon) + O2
 CO2 + H2O + energy
CH4 + 2O2  CO2 + 2H2O + 890kJ
∆H = -890kJ
Neutralization
Exothermic reaction
Acid + Base  Salt + Water + energy
HCl + NaOH  NaCl + H2O + 57.3 kJ
∆H = -57.3kJ
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6.5
Factors Affecting Heat Quantities
Most of these energy transfers involve heat
transfers and can be measured by
temperature changes.
 The amount of heat contained by an object
depends primarily on three factors:
 The mass of material
 The temperature
 The kind of material and its ability to absorb
or retain heat.

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Heat Quantities

The heat required to raise the temperature of
1.00 g of water 1 oC is known as a calorie
 Calorie (with a capital “C”): dietary
measurement of heat. Food has potential
energy stored in the chemical bonds of food.
1 Cal = 1 kcal = 1000 cal
 The SI unit for heat is the joule. It is based on
the mechanical energy requirements.
 1.00 calorie = 4.184 Joules
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Specific Heat
Specific heat: The amount of heat energy
required to raise the temperature of 1 g of a
substance by 1 oC.
Units are: J g-1 oC-1
Specific heat of water (or aqueous solutions)
= 4.184 Joules = 1 cal
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5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
Heat Energy Change
q = m x c x ΔT
ΔH = q = heat (joules)
m = mass (g)
c = specific heat (J g-1 oC-1)
The amount of heat required to raise the temperature of 1
g of a substance by 1 oC.
Specific heat of water = 4.184 Joules
ΔT = change in temperature
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5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
1. How much heat in joules 2. When 435 J of heat is
will be absorbed when
added to 3.4 g of olive oil
32.0 g of water is
at 21 oC, the temperature
heated from 25.0 oC to
increases to 85 oC. What is
80.0 oC?
the specific heat of olive
oil?
40
5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
How much heat in joules will be
absorbed when 32.0 g of
water is heated from 25.0 oC
to 80.0 oC?
When 435 J of heat is added to 3.4 g of
olive oil at 21 oC, the temperature
increases to 85 oC. What is the
specific heat of olive oil?
q = m x c x ΔT
q=?
m = 32.0 g
c = 4.18 J g-1 oC-1
ΔT = 80.0-25.0 = 55.0 oC
q = m x c x ΔT
q = 435 J
m = 3.4 g
c=?
ΔT = 85-21 = 64 oC
q = 32.0 x 4.18 x 55.0 = 7,360
J
435 = 3.4 x c x 64 = 2.0 J g-1 oC-1
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Heat Transfer Problem 3
Calculate the heat gained in an aluminum cooking pan
whose mass is 400 grams, from 20oC to 200oC. The
specific heat of aluminum is 0.902 J g-1 oC-1.
Solution
q = mCT
= (400 g) (0.902 J g-1 oC-1)(200oC – 20oC)
= 64,944 J = 60,000J OR 6 x 104 J OR 60 kJ
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5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
4. How much heat in joules 5. A 1.55 g piece of stainless
is required to raise the
steel absorbs 141 J of heat
temperature of 250 g of
when its temperature
mercury 52oC? The
increases to 178 oC. The
specific heat capacity of
specific heat of steel is
Hg is 0.14 J g-1 oC-1
0.511 J g-1 oC-1. What is
the initial temperature of
the steel?
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5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
4. How much heat in joules is
required to raise the
temperature of 250 g of
mercury 52 oC?
q = m x c x ΔT
q=?
m = 250 g
c = 0.14 J g-1 oC-1
ΔT = 52 oC
q = 250 x 0.14 x 52 = 1800 J
OR 1.8 kJ
5. A 1.55 g piece of stainless steel
absorbs 141 J of heat when its
temperature increases to 178 oC. The
specific heat of steel is 0.511 J g-1 oC-1.
What is the initial temperature of the
steel?
q = m x c x ΔT
q = 141 J
m = 1.55 g
c = 0.511 J g-1 oC-1
ΔT = 178 oC – x
141 = 1.55 x 0.511 x (178 – x)
= 0.00 oC
44
5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
Heat Energy Change
ΔH =
m x c x ΔT
mole x 1000
ΔH = enthalpy (kJ mol-1)
m = mass (g)
c = specific heat (J g-1 oC-1)
The amount of heat required to raise the temperature of 1
g of a substance by 1 oC.
Specific heat of water = 4.184 Joules
ΔT = change in temperature
45
5.2.2 Design suitable experimental procedures for
measuring the heat energy changes of reactions.

Calorimeter: Reactions
used to heat up an
external source of
water.
Temperature change of
water, mass of material
and mass of water are
measured.
Use q = m x c x ΔT to
solve for q then find the
heat of reaction in
kJ/mol of reacting
substance.
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Calorimetry

Calorimetry involves the measurement of
heat changes that occur in chemical
processes or reactions. Determines
the ΔH by measuring temp Δ's created
from the rxn

The heat change that occurs when a
substance absorbs or releases energy is
really a function of three quantities:



The mass
The temperature change
The heat capacity of the material
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47
Heat Capacity and Specific Heat



The ability of a substance to absorb or retain heat
varies widely.
The heat capacity depends on the nature of the
material.
The specific heat of a material is the amount of
heat required to raise the temperature of 1 gram of
a substance 1 oC (or Kelvin)
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Specific Heat values for Some
Common Substances
Substance
CJ g-1 K-1
C J mol-1K-1
Water (liquid)
4.184
75.327
Water (steam)
2.080
37.47
Water (ice)
2.050
38.09
Copper
0.385
24.47
Aluminum
0.897
24.2
Ethanol
2.44
112
Lead
0.127
26.4
49
5.2.2 Design suitable experimental procedures for
measuring the heat energy changes of reactions.
Heat Energy Change for Physical Processes
(mixing a hot substance with a cold substance)
q = -q
HUH?!?
50
5.2.2 Design suitable experimental procedures for
measuring the heat energy changes of reactions.
Heat Energy Change
qcold = -qhot
mcT= -mcT
the heat gained by the cooler substance
= the heat lost by the warmer substance
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Heat Transfer Problem 2
What mass of water will be heated from 5.72oC to
20.65oC if it is mixed in a calorimeter with 80.72 g of
water that starts at 29.5oC? Assume that the loss of
heat to the surroundings is negligible. The specific heat
of water is 4.184 J g-1 oC-1
Solution: Q (Cold) = -Q (hot)
mcT= -mcT
(m)(4.184)(20.65 – 5.72) = – (80.72)(4.184)(20.65 – 29.5)
(m)(4.184)(14.93) = – (80.72)(4.184)(–8.85)
62.46712 (m) = 2988.932448
m = 47.84 g
Using proper sig. fig.s…
m = 47.8 g
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Heat Transfer Problem 2
What mass of water will be heated from 5.72oC to
20.65oC if it is mixed in a calorimeter with 80.72 g of
water that starts at 29.5oC? Assume that the loss of
heat to the surroundings is negligible. The specific heat
of water is 4.184 J g-1 oC-1
Solution: Q (Cold) = -Q (hot)
mcT= -mcT
(m)(4.184)(20.65 – 5.72) = – (80.72)(4.184)(20.65 – 29.5)
(m)(4.184)(14.93) = – (80.72)(4.184)(–8.85)
62.46712 (m) = 2988.932448
m = 47.84 g
Using proper sig. fig.s…
m = 47.8 g
53
53
Heat Transfer Problem 3
What mass of water will be heated from 25oC to 38oC
when 5.0 grams of 98oC copper is added to the water?
Assume that heat loss to the surroundings is negligible.
Substance
C (J g-1 K-1)
C (J mol-1K-1)
Copper
0.385
24.47
Aluminum
0.897
24.2
Lead
0.127
26.4
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Heat Transfer Problem 3
What mass of water that will be heated from 25oC to
38oC when 5.0 grams of 98oC copper is added to the
water? Assume that the loss of heat to the
surroundings is negligible
Solution: Q (Cold: H2O) = -Q (hot Cu)
mcT= -mcT
Let T = final temperature
(x) (4.184)(38 - 25oC) = - (5.0g) (0.385)(38 - 98)
(x)(4.184)(13) = - (5.0 g)(0.385)(-60)
54.392x = 115.5
x = 2.123474 g
With proper sig. figs. x = 2.1 g
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Heat Transfer Problem 4
What is the final temperature when 50 grams of water at
20oC is added to 80 grams water at 60oC? Assume
that the loss of heat to the surroundings is negligible.
The specific heat of water is 4.184 J g-1 oC-1
Solution: Q (Cold) = -Q (hot)
mcT= -mcT
Let T = final temperature
(50 g) (4.184 J g-1 oC-1)(T – 20oC)
= –(80 g) (4.184 J g-1 oC-1)(T – 60oC)
(50 g)(T – 20oC) = –(80 g) (T – 60oC)
50T – 1,000 = 4,800 – 80T
130T = 5,800
T = 44.6 oC
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5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Thermochemical Equations
Calcium oxide combines with water to produce calcium hydroxide and heat
(exothermic reaction).

CaO(s) + H2O(l)  Ca(OH)2(s) + 65.2 kJ OR

CaO(s) + H2O(l)  Ca(OH)2(s)
H = -65.2 kJ

These H values assume 1 mole of each compound (based on the coefficients)
How many kJ of heat are produced when 7.23 g of CaO react?
1)
Write out and balance equation:
2)
Determine the number of moles:
3)
Multiply:
* Notice that mol / mol cancel out and you’re left with kJ
4)
Solve =
57
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Thermochemical Equations
Calcium oxide combines with water to produce calcium hydroxide and heat
(exothermic reaction).

CaO(s) + H2O(l)  Ca(OH)2(s) + 65.2 kJ OR

CaO(s) + H2O(l)  Ca(OH)2(s)
H = -65.2 kJ

These H values assume 1 mole of each compound (based on the coefficients)
How many kJ of heat are produced when 7.23 g of CaO react?
1)
Write out and balance equation: already balanced
2)
Determine the number of moles: 7.23 g/56.01 gmol-1 = 0.129 mol
3)
Multiply: 0.129 mol CaO x 65.2 kJ mol-1
* Notice that mol / mol cancel out and you’re left with kJ
4)
Solve = 8.41 kJ
58
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Thermochemical Equations
Sodium hydrogen carbonate absorbs 129 kJ of energy and decomposes to
sodium carbonate, water, and carbon dioxide.
How many kJ of heat are needed to decompose 2.24 mol NaHCO3(s)?
1)
Write out and balance the equation
2)
Determine the number of moles of NaHCO3(s)
3)
Set up the ratio
4)
Solve for x
5)
State, with justification, whether the reaction is endothermic or
exothermic.
59
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Thermochemical Equations
Sodium hydrogen carbonate absorbs 129 kJ of energy and decomposes to
sodium carbonate, water, and carbon dioxide.

2NaHCO3(s) + 129 kJ  Na2CO3(s) + H2O(g) + CO2(g) OR

2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g) H = 129 kJ
How many kJ of heat are needed to decompose 2.24 mol NaHCO3(s)?
1)
Balance equation
2)
Moles NaHCO3(s) = 2.24 mol
3)
Ratio:
x kJ
=
129 kJ
2.24 NaHCO3(s)
2 mol
4)
Solve for x = 144 kJ
5)
The reaction is endothermic because energy is being absorbed / the
H is positive.
60
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Experimental Data
In a neutralization reaction, 25.0 mL of water containing 0.025 mol HCl is
added to 25.0 mL of water containing 0.025 mol NaOH in a foam cup
calorimeter. At the start, the solutions and the calorimeter are all at 25.0
oC. During the reaction, the highest temperature observed is 32.0 oC.
Calculate the heat (in kJ) released during this reaction. Assume the
densities of the solutions are 1.00 g mL-1.
61
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Experimental Data
In a neutralization reaction, 25.0 mL of water containing 0.025 mol HCl is
added to 25.0 mL of water containing 0.025 mol NaOH in a foam cup
calorimeter. At the start, the solutions and the calorimeter are all at 25.0
oC. During the reaction, the highest temperature observed is 32.0 oC.
Calculate the heat (in kJ) released during this reaction. Assume the
densities of the solutions are 1.00 g mL-1.
Use q = m x c x ΔT
m = mass of solution = 50.0 mL x 1.00 g mL-1 = 50.0 g
C = 4.18 j g-1 oC-1
ΔT = 32.0 – 25.0 = 7.0 oC
q = 50.0 g x 4.18 J g-1 oC-1 x 7.0 oC = 1463 J = 1.5 kJ
62
IB Topic 5: Energetics
5.2: Hess’s Law

The enthalpy change for a reaction that is
carried out in a series of steps is equal to the
sum of the enthalpy changes for the individual
steps.
63
5.3.1 Determine the enthalpy change of a reaction that
is the sum of two or three reactions with known
enthalpy changes.
Hess’s Law: Reactions can be added together in order to
determine heats of reactions that can’t be measured in
the lab.
The enthalpy change of a reaction is the sum of two or
three reactions with known enthalpy changes.
64
5.3.1 Determine the enthalpy change of a reaction that
is the sum of two or three reactions with known
enthalpy changes.
The enthalpy change of a reaction is the sum of
two or three reactions with known enthalpy
changes.
Rules for applying Hess’s Law:
1.
2.
3.
Make sure to rearrange the given equations so that
reactants and products are on the appropriate sides of
the arrows.
If you reverse equations, you must also reverse the
sign of ΔH.
If you multiply equations to obtain a correct coefficient,
you must also multiply the ΔH by this coefficient.
65
5.3.1 Determine the enthalpy change of a reaction that
is the sum of two or three reactions with known
enthalpy changes.
C(diamond)  C(graphite)
This reaction is too slow to be measured in the lab. Two reactions can
be used that can be measured in the lab:
a) C(graph) + O2(g)  CO2(g)
H = -393.5 kJ
b) C(diam) + O2(g)  CO2(g)
H = -395.4 kJ
Since C(graphite) is a product, write equation a) in reverse to give:
c) CO2(g)  C(graph) + O2(g) H = 393.5 kJ
Now add equations b) and c) together:
C(diam) + O2(g) + CO2(g)  C(graph) + O2(g) + CO2(g)
H = -395.4 kJ + 393.5 kJ = -1.9 kJ
Final equation: C(diamond)  C(graphite) H = - 1.9 kJ
66
5.3.1 Determine the enthalpy change of a reaction that
is the sum of two or three reactions with known
enthalpy changes.
Given the following thermochemical equations, calculate the heat of reaction
for:
C2H4(g) + H2O(l)  C2H5OH(l)
a) C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
b) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l)
H = -1367 kJ
H = -1411 kJ
67
5.3.1 Determine the enthalpy change of a reaction that
is the sum of two or three reactions with known
enthalpy changes.
Given the following thermochemical equations, calculate the heat of reaction
for
C2H4(g) + H2O(l)  C2H5OH(l)
a) C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
b) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l)
H = -1367 kJ
H = -1411 kJ
Since C2H5OH(l) is a product, write equation a) in reverse order:
c) 2CO2(g) + 2H2O(l)  C2H5OH(l) + 3O2(g)
H = 1367 kJ
Add equations b) & c) together, cancelling out substances on opposite sides of
the arrow. Add the heat values to obtain the heat of reaction.
C2H4(g) + H2O(l)  C2H5OH(l)
H = -44 kJ
68
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Thermochemical Equations
Mg(s) + ½ O2(g)  MgO(s) + heat

Mg(s) + 2HCl(aq)
 MgO(s) + 2HCl(aq)
 H2(g) + ½ O2(g)
Equation 1
 MgCl2(aq) + H2(g) Equation A
 MgCl2(aq) +H2O(l) Equation B
 H2O(l)
Equation C
∆H = -A
∆H = -B
∆H = +C
69
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Thermochemical Equations
Mg(s) + ½ O2(g)  MgO(s) + heat

Mg(s) + 2HCl(aq)
 MgO(s) + 2HCl(aq)
 H2(g) + ½ O2(g)

Equation 1
 MgCl2(aq) + H2(g) Equation A
 MgCl2(aq) +H2O(l) Equation B
 H2O(l)
Equation C
∆H = -A
∆H = -B
∆H = +C
The final ∆H = -A + B + C
70
5.3.1 Determine the enthalpy change of a reaction that
is the sum of two or three reactions with known
enthalpy changes.
Given the following thermochemical equations, calculate the heat of reaction
for:
C(s) + 2H2(g)  CH4(g)
a) C(s) + O2(g)  CO2(g)
b) H2(g) + ½ O2(g)  H2O(l)
c) CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H = -393 kJ mol-1
H = -286 kJ mol-1
H = -890 kJ mol-1
71
5.3.1 Determine the enthalpy change of a reaction that
is the sum of two or three reactions with known
enthalpy changes.
Given the following thermochemical equations, calculate the heat of reaction
for:
C(s) + 2H2(g)  CH4(g)
C(s) + O2(g)  CO2(g)
H = -393 kJ
2(H2(g) + ½ O2(g)  H2O(l))
2H2(g) + O2(g)  2H2O(l))
H = 2(-286 kJ mol-1)
H = -572 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
CO2(g) + 2H2O(l)  CH4(g) + 2O2(g)
H = -890 kJ mol-1
H = 890
C(s) + 2H2(g)  CH4(g)
H = -75 kJ
72
Energy Changes

Breaking chemical bonds requires energy
 Endothermic

Forming new chemical bonds releases
energy
 Exothermic
73
73
Enthalpy Change of Formation

Enthalpy change of formation (Hf) is the
enthalpy that accompanies the formation
of 1 mole of compound from it’s
constituent elements in their gaseous form
74
74
Energy Changes

Ex. Write the reaction for the enthalpy
change of formation of one mole of
ethanol, C2H5OH.
Hf = C (g) + H2 (g) + O2 (g)  C2H5OH (l)
 Now balance…
2C (g) + 3H2 (g) + ½O2 (g)  C2H5OH
75
75
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Thermodynamic Quantities (Standard Heats of Formation)

Heat of reaction can be found by: sum the heats of formation of all
the products – sum of heats of formation of all the reactants
Hrxn = Hf products – Hf reactants



Hf = standard enthalpy of formation. Energy required to form a
compound from its elements.
“standard” is a term used a lot in chemistry. It usually means that
the values are experimentally determined and compared to an
agreed upon reference value
Since the Hf is given per mole, we must multiply by coefficients
76
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Using Thermodynamic Quantities (Standard Heats of
Formation)
Using the table of thermodynamic quantities, calculate the heat of
reaction for 2SO2(g) + O2(g)  2SO3(g)
Heat of reaction = Hf products – Hf of reactants
Heat of products: Hf (SO3) = -395.2 kJ/mol x 2 = -790.4 kJ
Heat of reactants = Hf (SO2) + Hf (O2)
(-296.9 kJ/mol x 2) + (0) = -593.8 KJ
Heat of reaction = -790.4 kJ – (-593.8 kJ) = -196.6 kJ
77
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
N2O4 + 3 CO  N2O + 3CO2
Reactants
Hf
Products
Hf
N 2 O4
9.7 kJ mol-1
N 2O
81 kJ mol-1
CO
-110 kJ mol-1
CO2
-393 kJ mol-1
78
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
N2O4 + 3 CO  N2O + 3CO2
Reactants
Hf
Products
Hf
N 2 O4
9.7 kJ mol-1
N 2O
81 kJ mol-1
CO
-110 kJ mol-1
CO2
-393 kJ mol-1
Hf products = 1(81 kJ mol-1) + 3(-393 kJ mol-1) = -1098 kJ/mol
Hf reactants = 1(-9.7 kJ mol-1)+ 3(-110 kJ mol-1) = -320.3 kJ mol-1
Hf products – Hf reactants = (-1098 kJ mol-1) – (-320.3 kJ mol-1)
= -778 kJ mol-1
Hrxn = -778 kJ mol-1
Therefore it is exothermic
79
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Ca(OH)2(s) + CO2 (g)  H2O(g) + CaCO3 (s)
Reactants
Hf
Products
Hf
Ca(OH)2
-986.1 kJ mol-1
H 2O
-241.8 kJ mol-1
CO2
-393.5 kJ mol-1
CaCO3
-1206.9 kJ mol-1
80
5.2.3 Calculate the enthalpy change for a reaction using
experimental data on temperature changes, quantities of
reactants and mass of water.
Ca(OH)2(s) + CO2 (g)  H2O(g) + CaCO3 (s)
Reactants
Hf
Products
Hf
Ca(OH)2
-986.1 kJ mol-1
H 2O
-241.8 kJ mol-1
CO2
-393.5 kJ mol-1
CaCO3
-1206.9 kJ mol-1
Hf products = 1(-241.8 kJ mol-1) + 1(-1206.9 kJ mol-1) = -1448.7 kJ/mol
Hf reactants = 1(-986.1 kJ mol-1)+ 1(-393.5 kJ mol-1) = -1379.6 kJ mol-1
Hf products – Hf reactants = (-1448.7 kJ mol-1) – (-1379.6 kJ mol-1)
= -69.1 kJ mol-1
Hrxn = -69.1 kJ mol-1
Therefore it is exothermic
81
Methods of Calculating Enthalpy
There are four ways of calculating ΔH for a
reaction: by using…
1. bond enthalpies
2. experimental data to calculate q
3. enthalpy of formation values
4. Hess’s Law
82
Methods of Calculating Enthalpy
There are four ways of calculating ΔH for a
reaction: by using…
1. bond enthalpies
2. experimental data to calculate q

3.
4.
Then dividing by 1000 and moles to get kJ
mol-1
enthalpy of formation values
Hess’s Law
83
Using Bond Enthalpies
H =
∑ (bond enthalpy
of reactants)
–
∑(bond enthalpy
of products)
84
Using Experimental Data
Heat Energy Change
ΔH =
m x c x ΔT
mole x 1000
ΔH = enthalpy (kJ mol-1)
m = mass (g)
c = specific heat (J g-1 oC-1)
The amount of heat required to raise the temperature of 1
g of a substance by 1 oC.
Specific heat of water = 4.184 Joules
ΔT = change in temperature
85
Using Heat of Formation

Heat of reaction can be found by: sum the heats of
formation of all the products – sum of heats of
formation of all the reactants
Hrxn = Hf products – Hf reactants

Hf = standard enthalpy of formation = enthalpy that
accompanies the formation of 1 mole of compound
from it’s constituent elements in their gaseous form.
86
Using Hess’s Law
The enthalpy change of a reaction is the sum of
two or three reactions with known enthalpy
changes.
Rules for applying Hess’s Law:
1.
2.
3.
4.
Rearrange given equations Rs & Ps match unknown.
If equations reverse, reverse ΔH sign.
If coefficient changes, multiply the ΔH by coefficient.
Add ΔH of known values.
87
Terms to Know











Endothermic
Exothermic
Temperature
Heat
Enthalpy
Standard enthalpy change of reaction
Standard enthalpy of formation
Enthalpy of combustion
Average bond enthalpy
Hess’ Law
Standard conditions
88